Answer
\[A=4\]
Work Step by Step
\[\begin{align}
& \text{From the graph} \\
& R=\left\{ \left( x,y \right):1-\sin x\le y\le 1+\sin x,\text{ }0\le x\le \pi \text{ } \right\} \\
& \text{Then,} \\
& A=\int_{0}^{\pi }{\int_{1-\sin x}^{1+\sin x}{dy}dx} \\
& \text{Integrate} \\
& A=\int_{0}^{\pi }{\left[ y \right]_{1-\sin x}^{1+\sin x}dx} \\
& A=\int_{0}^{\pi }{\left[ \left( 1+\sin x \right)-\left( 1-\sin x \right) \right]dx} \\
& A=\int_{0}^{\pi }{\left( 1+\sin x-1+\sin x \right)}dx \\
& A=\int_{0}^{\pi }{2\sin x}dx \\
& A=-2\left[ \cos x \right]_{0}^{\pi } \\
& A=-2\left[ \cos \pi -\cos 0 \right] \\
& A=-2\left( -2 \right) \\
& A=4 \\
\end{align}\]
